Question

Determinati in fiecare dintr urmatoarele cazuri S a solutiilor complexe a urmatoarelor ecuatii : 
a. x²-3x+2=0
b. x²-4c+5=0
c. x²+6x+10=0
d. 2x²-3x+2=0
e. x²-2x+5=0
f. x²-6x+13=0

Answer 1

a)$x ^{2} -3x+2=0$

Δ$= b^{2} -4ac=(-3) ^{2} -4*1*2=9-8=1$

$x_{1} = \frac{-b- \sqrt{delta} }{2a} = \frac{3-1}{2} = \frac{2}{2} =1$

$x_{2} = \frac{-b+ \sqrt{delta} }{2a} = \frac{3+1}{2} = \frac{4}{2} =2$

S={1;2}

b) $x ^{2} -4x+5=0$

Δ$= b^{2} -4ac=(-4) ^{2} -4*1*5=16-20=-4$

$x_{1}= \frac{-b-i \sqrt{-delta} }{2a} = \frac{4-i \sqrt{4} }{2} = \frac{4-2i}{2} = \frac{2(2-i)} {2}=2-i$

$x_{2}= \frac{-b+i \sqrt{-delta} }{2a} = \frac{4+i \sqrt{4} }{2} = \frac{4+2i}{2} = \frac{2(2+i)}{2}=2+i$

S={2-i;2+i}

c)$x^{2} +6x+10=0$

Δ$= b^{2} -4ac=6 ^{2} -4*1*10=36-40=-4$

$x_{1}= \frac{-b-i \sqrt{-delta} }{2a} = \frac{-6-i \sqrt{4} }{2} = \frac{6-2i}{2} = \frac{2(3-i)}{2}=3-i$

$x_{2}= \frac{-b+i \sqrt{-delta} }{2a} = \frac{-6+i \sqrt{4} }{2} = \frac{6+2i}{2} = \frac{2(3+i)}{2}=3+i$

S={3-i;3+i}

d)$2x^{2} -3x+2=0$

Δ$= b^{2} -4ac=(-3) ^{2} -4*2*2=9-16=-7$

$x_{1}= \frac{-b-i \sqrt{-delta} }{2a} = \frac{3-i \sqrt{7} }{4}$

$x_{2}= \frac{-b+i \sqrt{-delta} }{2a} = \frac{3+i \sqrt{7} }{4}$

S={$\frac{3-i \sqrt{7} }{4}$;$\frac{3+i \sqrt{7} }{4}$}

e)$x^{2} -2x+5=0$

Δ$= b^{2} -4ac=(-2) ^{2} -4*1*5=4-20=-16$

$x_{1}= \frac{-b-i \sqrt{-delta} }{2a} = \frac{2-i \sqrt{16} }{2} = \frac{2-4i}{2} = \frac{2(1-2i)}{2}=1-2i$

$x_{2}= \frac{-b-i \sqrt{-delta} }{2a} = \frac{2+i \sqrt{16} }{2} = \frac{2+4i}{2} = \frac{2(1+2i)}{2}=1+2i$

S={1-2i;1+2i}

f)$x^{2} -6x+13=0$

Δ$= b^{2} -4ac=(-6) ^{2} -4*1*13=36-52=-16$

$x_{1}= \frac{-b-i \sqrt{-delta} }{2a} = \frac{6-i \sqrt{16} }{2} = \frac{6-4i}{2} = \frac{2(3-2i)}{2}=3-2i$

$x_{1}= \frac{-b+i \sqrt{-delta} }{2a} = \frac{6+i \sqrt{16} }{2} = \frac{6+4i}{2} = \frac{2(3+2i)}{2}=3+2i$

S={3-2i;3+2i}